ABC is an isosceles triangle where AB = AC which is circumscribed about a circle. If P is the point where the circle touches the side BC, then which of the following is true?
A. BP = PC
B. BP > PC
C. BP < PC
D. BP = $$\frac{1}{2}$$PC
Answer: Option A
Solution (By Examveda Team)

Here given that = AB = AC
AQ + BQ = AR + RC
We know that
BQ = PB & PC = RC
AQ + PB = AR + PC
Also AQ = AR
AR + PB = AR + PC
PB = PC
Related Questions on Geometry
A. $$\frac{{23\sqrt {21} }}{4}$$
B. $$\frac{{15\sqrt {21} }}{4}$$
C. $$\frac{{17\sqrt {21} }}{5}$$
D. $$\frac{{23\sqrt {21} }}{5}$$
In the given figure, ∠ONY = 50° and ∠OMY = 15°. Then the value of the ∠MON is

A. 30°
B. 40°
C. 20°
D. 70°


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